Longest Palindromic Substring
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public class Solution { public String longestPalindrome(String s) { char[] cc=s.toCharArray(); int[][] dp=new int[cc.length+1][cc.length+1]; int maxlen=0; int maxstart=0; int maxend=0; for(int i=cc.length-1;i>=0;i--) for(int j=i;j<cc.length;j++) if(cc[i]==cc[j]&&(j-i<2||dp[i+1][j-1]==1)){ dp[i][j]=1; if(maxlen<j-i){ maxstart=i; maxend=j; maxlen=j-i; } } return s.substring(maxstart,maxend+1); }}
上为仿照palindrome partition 代码版本,下为bottom up 版本速度比上面的快一倍
public String longestPalindrome(String s) { char[] cc=s.toCharArray(); String res=""; int maxlen=-1; for(int i=0;i<cc.length*2;i++){ if(i/2-maxlen<0||i/2+maxlen>=cc.length) continue; int j=0; if(i%2==1) j=1; for(;j<=cc.length;j=j+2){ if(i-j<0||(i+j)/2>=cc.length) break; if(cc[(i-j)/2]!=cc[(i+j)/2]) break; if(maxlen*2<j){ maxlen=j/2; res=s.substring((i-j)/2,(i+j)/2+1); } } } return res; }
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